This question may be very trivial in this area, but I am a beginner to this and I stuck here. Could anyone please help me here!

Let $\Gamma$ be a group whose identity is $e$.

Let $X$ be a set and $∗:\Gamma×S\rightarrow S$ be a group action.

Let $[x]$ is the orbit (equivalence class) of $x\in X$ under the group action $\Gamma$.

Now I want to define addition in the quotient space $X/\Gamma$.

Here is my attempt:

Take $x'\in [x]$ and $y'\in [y]$ where $[x]\cap[y]=\phi$.

Then to define sum uniquely, I need to show $x+x'\sim_G y+y'$.

To show the above: Since $x'\in [x]$ and $y'\in [y]$ $$\exists g_1\in \Gamma: x'=g_1*x, \;\;\exists g_2\in \Gamma: y'=g_2*y.$$

Now $$x'+y'=g_1*x+g_2*y$$ Here I stuck: since $g_1$ and $g_2$ may be different element of $\Gamma$, then how to show $x'+y'\sim_\Gamma x+y$?

  • 2
    $\begingroup$ But you haven't defined $x+y$ and unitl you do you can't prove anything about it! I don't believe that there is any natural way of doing this that works in complete generality for any group action. You would normally only want to do it if you already had an additive group structure on $S$. $\endgroup$ – Derek Holt Jan 23 '16 at 10:03
  • $\begingroup$ @DerekHolt I got your point...thank you. $\endgroup$ – Janak Feb 5 '16 at 6:49

When you have an action of a group to a set then the quotient doesn't always form a group. Take for example the natural action of the group of orthogonal matrices $O(n)$ on the sphere $\mathbb{S}^{n-1}$. Let $p$ be the north pole, i.e. $p=(0,0,\cdots,0,1)$, then you have that $$O(n)/\mathrm{Stab}(p)\cong \mathbb{S}^{n-1},$$ where $\cong$ means diffeomorphism and $\mathrm{Stab}(p)$ is the stabilizer of $p$ in $O(n)$ (which is isomophic to $O(n-1)$). Thus, if the quotient $O(n)/\mathrm{Stab}(p)$ could form a group then we could pass that group structure to $\mathbb{S}^{n-1}$ via the diffeomorphism. But this is a contradiction since (due to J.Milnor) we know that the only spheres that admit a group structure are $\mathbb{S}^1$ and $\mathbb{S}^3$.

  • $\begingroup$ $S^1$ and $S^3$ are the only spheres admitting a continuous group structure in the natural topology. So I don't think this example really holds water. $\endgroup$ – rogerl Jan 23 '16 at 13:21
  • $\begingroup$ @rogerl didnt know that.. Anyway let $M$ be a complete Riemannian manifold with fundamental group $\pi_1(M)$ and $\tilde{M}$ its universal covering. Then we know that $M$ can be viewed as the quotient space $\tilde{M}/\Gamma$, where $\Gamma\cong \pi_1(M)$ is a discrete group of isometries of $\tilde{M}$ acting freely on $\tilde{M}$. Therefore the quotient doesnt always form a group. $\endgroup$ – Christos Jan 23 '16 at 13:45
  • $\begingroup$ But again what you're saying is that the operation induced on the quotient does not produce a group structure on the quotient. Your answer starts "When you have an action of a group to a set then the quotient doesn't always form a group", but if you simply regard the quotient space as a set, of course you can define a group structure on it - for example, it is set-isomorphic to $\mathbb{C}$. So while I understand what you are trying to say here, you haven't said it precisely... $\endgroup$ – rogerl Jan 23 '16 at 13:48

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